Mathematical Tools for Physics

5—Fourier Series 125

No solutions there, so tryλ= 0

u(x) =A+Bx, then u(0) =A= 0 and so u(L) =BL= 0⇒B= 0

No solutions here either. Tryλ < 0 , settingλ=−k^2.

u(x) =Asinkx+Bcoskx, then u(0) = 0 =B and so u(L) =AsinkL= 0

Now there are many solutions becausesinnπ= 0allowsk=nπ/Lwithnany integer. But,sin(−x) =−sin(x)
so negative integers just reproduce the same functions as do the positive integers; they are redundant and you
can eliminate them. The complete set of solutions to the equationu′′=λuwith these boundary conditions have
λn=−n^2 π^2 /L^2 with

un(x) = sin

(nπx

L

)

n= 1, 2 , 3 ,... and

〈

un,um

〉

=

∫L

0

dxsin

(nπx

L

)

sin

(mπx

L

)

= 0 if n 6 =m (14)

There are other choices of boundary condition that will make the bilinear concomitant vanish. For example

u(0) = 0, u′(L) = 0 gives un(x) = sin

(

n+^1 / 2

)

πx/L n= 0, 1 , 2 , 3 ,...

and you have the orthogonality integral for non-negative integersnandm

∫L

0

dxsin

(

(n+^1 / 2 )πx

L

)

sin

(

(m+^1 / 2 )πx

L

)

= 0 if n 6 =m (15)

A very common choice of boundary conditions is

u(a) =u(b), u′(a) =u′(b) (periodic boundary conditions) (16)

It’s often more convenient to use complex exponentials in this case (though of course not necessary). On
0 < x < L
u(x) =eikx, where k^2 =−λ and u(0) = 1 =u(L) =eikL